WEEK 9

1. Jordan-Holder¨ theorem and indecomposable modules Let M be a satisfying ascending and descending chain conditions (ACC and DCC). In other words every increasing sequence submodules M1 ⊂ M2 ⊂ ... and any decreasing sequence M1 ⊃ M2 ⊃ ... are finite. Then it is easy to see that there exists a finite sequence

M = M0 ⊃ M1 ⊃···⊃ Mk = 0 such that Mi/Mi+1 is a . Such a sequence is called a Jordan-H¨older series. We say that two Jordan H¨older series

M = M0 ⊃ M1 ⊃···⊃ Mk = 0, M = N0 ⊃ N1 ⊃···⊃ Nl = 0 ∼ are equivalent if k = l and for some permutation s Mi/Mi+1 = Ns(i)/Ns(i)+1. Theorem 1.1. Any two Jordan-H¨older series are equivalent. Proof. We will prove that if the statement is true for any submodule of M then it is true for M. (If M is simple, the statement is trivial.) If M1 = N1, then the ∼ statement is obvious. Otherwise, M1 + N1 = M, hence M/M1 = N1/ (M1 ∩ N1) and ∼ M/N1 = M1/ (M1 ∩ N1). Consider the series

M = M0 ⊃ M1 ⊃ M1∩N1 ⊃ K1 ⊃···⊃ Ks = 0, M = N0 ⊃ N1 ⊃ N1∩M1 ⊃ K1 ⊃···⊃ Ks = 0. They are obviously equivalent, and by induction assumption the first series is equiv- alent to M = M0 ⊃ M1 ⊃ ··· ⊃ Mk = 0, and the second one is equivalent to M = N0 ⊃ N1 ⊃···⊃ Nl = {0}. Hence they are equivalent.  Thus, we can define a length l (M) of a module M satisfying ACC and DCC, and if M is a proper submodule of N, then l (M)

Lemma 1.2. Let M and N be indecomposable, α ∈ HomR (M, N), β ∈ HomR (N, M) be such that β ◦ α is an isomorphism. Then α and β are isomorphisms. Proof. We claim that N = Im α⊕Ker β. Indeed, Im α∩Ker β = 0 and for any x ∈ N one can write x = y + z, where y = α ◦ (β ◦ α)−1 ◦ β (x), z = x − y. Then since N is indecomposable, Im α = N, Ker β = 0 and N =∼ M. 

Date: November 7, 2005. 1 2 REPRESENTATION THEORY WEEK 9

Lemma 1.3. Let M be indecomposable module of finite length and ϕ ∈ EndR (M), then either ϕ is an isomorphism or ϕ is nilpotent. Proof. There is n > 0 such that Ker ϕn = Ker ϕn+1, Im ϕn = Im ϕn+1. In this case Ker ϕn ∩ Im ϕn = 0 and hence M ∼= Ker ϕn ⊕ Im ϕn. Either Ker ϕn = 0, Im ϕn = M or Ker ϕn = M. Hence the lemma. 

Lemma 1.4. Let M be as in Lemma 1.3 and ϕ,ϕ1,ϕ2 ∈ EndR (M), ϕ = ϕ1 + ϕ2. If ϕ is an isomorphism then at least one of ϕ1, ϕ2 is also an isomorphism.

Proof. Without loss of generality we may assume that ϕ = id. But in this case ϕ1 and ϕ2 commute. If both ϕ1 and ϕ2 are nilpotent, then ϕ1 + ϕ2 is nilpotent, but this is impossible as ϕ1 + ϕ2 = id. 

Corollary 1.5. Let M be as in Lemma 1.3. Let ϕ = ϕ1 + ··· + ϕk ∈ EndR (M). If ϕ is an isomorphism then ϕi is an isomorphism at least for one i. It is obvious that if M satisfies ACC and DCC then M has a decomposition

M = M1 ⊕···⊕ Mk, where all Mi are indecomposable. Theorem 1.6. (Krull-Schmidt) Let M be a module of finite length and

M = M1 ⊕···⊕ Mk = N1 ⊕···⊕ Nl for some indecomposable Mi and Nj . Then k = l and there exists a permutation s ∼ such that Mi = Ns(j).

Proof. Let pi : M1 → Ni be the restriction to M1 of the natural projection M → Ni, and qj : Nj → M1 be the restriction to Nj of the natural projection M → M1. Then obviously q1p1 + ··· + qlpl = id, and by Corollary 1.5 there exists i such that qipi is an isomorphism. Lemma 1.2 implies that M1 =∼ Ni. Now one can easily finish the proof by induction on k. 

2. Some facts from homological algebra

The complex is the graded C· = ⊕i≥0Ci. We will assume later that all Ci are R-modules for some ring R. A differential is an R-morphism of degree −1 2 such that d = 0. Usually we realize C· by the picture d d −→· · · → C1 −→ C0 → 0. We also consider d of degree 1, in this case the superindex C· and d d 0 → C0 −→ C1 −→ ... All the proofs are similar for these two cases. Homology group is Hi (C) = (Ker d ∩ Ci) /dCi+1. REPRESENTATION THEORY WEEK 9 3

′ ′ ′ Given two complexes (C·, d) and (C· , d ). A morphism f : C· → C· preserving grading and satisfying f ◦ d = d′ ◦ f is called a morphism of complexes. A morphism ′ of complexes induces the morphism f∗ : H· (C) → H· (C ). Theorem 2.1. (Long exact sequence). Let

g ′ f ′′ 0 → C· −→ C· −→ C· → 0 be a short exact sequence, then the long exacts sequence

δ g∗ ′ f∗ ′′ δ g∗ −→ Hi (C) −→ Hi (C ) −→ Hi (C ) −→ Hi−1 (C) −→ ... where δ = g−1 ◦ d′ ◦ f −1, is exact. ′ Let f, g : C· → C· be two morphisms of complexes. We say that f and g are ′ homotopically equivalent if there exists h : C· → C· (+1) (the morphism of degree 1) such that f − g = h ◦ d + d′ ◦ h.

Lemma 2.2. If f and g are homotopically equivalent then f∗ = g∗.

Proof. Let φ = f − g, x ∈ Ci and dx = 0. Then φ (x)= h (dx)+ d′ (hx)= d′ (hx) ∈ Im d′.

Hence f∗ − g∗ = 0. 

′ We say that complexes C· and C· are homotopically equivalent if there exist f : ′ ′ C· → C· and g : C· → C· such that f ◦ g is homotopically equivalent to idC′ and g ◦ f is homotopically equivalent to idC . Lemma 2.2 implies that homotopically equivalent complexes have isomorphic homology. The following Lemma is straightforward.

′ Lemma 2.3. If C· and C· are homotopically equivalent then the complexes HomR (C·, B) ′ and HomR (C· ) are also homotopically equivalent.

Note that the differential in HomR (C·, B) has degree 1.

3. Projective modules An R-module P is projective if for any surjective morphism φ : M → N and any ψ : P → N there exists f : P → M such that ψ = φ ◦ f.

Example. A free module is projective. Indeed, let {ei}i∈Ibe the set of free generators of a free module F , i.e. F = ⊕i∈IRei. Define f : F → M by f (ei) = −1 φ (ψ (ei)). Lemma 3.1. The following conditions on a module P are equivalent (1) P is projective; (2) There exists a free module F such that F =∼ P ⊕ P ′; (3) Any exact sequence 0 → N → M → P → 0 splits. 4 REPRESENTATION THEORY WEEK 9

Proof. (1) ⇒ (3) Consider the exact sequence

ϕ ψ 0 → N −→ M −→ P → 0, then since ψ is surjective, there exists f : P → M such that ψ ◦ f = idP . (3) ⇒ (2) Every module is a quotient of a free module. Therefore we just have to apply (3) to the exact sequence 0 → N → F → P → 0 for a free module F . (2) ⇒ (1) Let φ : M → N be surjective and ψ : P → N. Choose a free module F so that F = P ⊕P ′. Then extend ψ to F → N in the obvious way and let f : F → M be such that φ◦f = ψ. Then the last identity is true for the restriction of f to P . 

A projective resolution of M is a complex P· of projective modules such that ∼ Hi (P·)=0for i> 0 and H0 (P·) = M. A projective resolution always exists since one can easily construct a resolution by free modules. Below we prove the “uniqueness” statement.

′ Lemma 3.2. Let P· and P· be two projective resolutions of the same module M. ′ Then there exists a morphism f : P· → P· of complexes such that f∗ : H0 (P·) → ′ H0 (P· ) induces the identity idM . Any two such morphisms f and g are homotopically equivalent.

′ ′ Proof. Construct f inductively. Let p : P0 → M and p : P0 → M be the natural ′ ′ projections, define f : P0 → P0 so that p ◦ f = p. Then ′ ′ ′ ′ f (Ker p) ⊂ Ker p , Ker p = d (P1), Ker p = d (P1) , ′ ′ ′ ′ hence f ◦ d (P1) ⊂ d (P1), and one can construct f : P1 → P1 such that f ◦ d = d ◦ f. Proceed in the same manner to construct f : Pi → Pi. To check the second statement, let ϕ = f − g. Then p′ ◦ ϕ = 0. Hence

′ ′ ′ ϕ (P0) ⊂ Ker p = d (P1) . ′ ′ Therefore one can find h : P0 → P1 such that d ◦ h = ϕ. Furthermore, d′ ◦ h ◦ d = ϕ ◦ d = d′ ◦ ϕ, hence ′ ′ ′ ′ (ϕ − h ◦ d)(P1) ⊂ P1 ∩ Ker d = d (P2) . ′ ′ Thus one can construct h : P1 → P2 such that d ◦ h = ϕ − h ◦ d. Then proceed ′  inductively to define h : Pi → Pi+1. Corollary 3.3. Every two projective resolutions of M are homotopically equivalent. REPRESENTATION THEORY WEEK 9 5

Let M and N be two modules and P· be a projective resolution of M. Consider the complex 0 → HomR (P0, N) → HomR (P1, N) → ..., where the differential is defined naturally. The cohomology of this complex is denoted · · by ExtR (M, N). Lemma 2.3 implies that ExtR (M, N) does not depend on a choice 0 of projective resolution for M. Check that ExtR (M, N) = HomR (M, N). Example 1. Let R = C [x] be the polynomial ring. Any simple R-module is one-dimensional and isomorphic to C [x] / (x − λ). Denote such module by Cλ.A projective resolution of Cλ is d 0 → C [x] −→ C [x] → 0, · where d (1) = x − λ. Let us calculate Ext (Cλ, Cµ). Note that HomC[x] (Cµ) = C, hence we have the complex d∗ 0 → C −→ C → 0 ∗ · 0 1 where d = λ−µ. HenceExt (Cλ, Cµ)=0if λ =6 µ and Ext (Cλ, Cλ) = Ext (Cλ, Cλ)= C. Example 2. Let R = C [x] / (x2). Then R has one up to isomorphism simple module, denote it by C0. A projective resolution for C0 is d d ... −→ R −→ R → 0, i where d (1) = x and Ext (C0, C0)= C for all i ≥ 0.

4. Representations of artinian rings An artinian ring is a unital ring satisfying the descending chain condition for left ideals. We will see that an artinian ring is a finite length module over itself. Therefore R is automatically noetherian. A typical example of an artinian ring is a finite-dimensional algebra over a field. Theorem 4.1. Let R be an artinian ring, I ⊂ R be a left ideal. If I is not nilpotent, then I contains an . Proof. Let J be a minimal left ideal, such that J ⊂ I and J is not nilpotent. Then J 2 = J. Let L be a minimal left ideal such that L ⊂ J and JL =6 0. Then there is x ∈ L such that Jx =6 0. But then Jx = L by minimality of L. Thus, for some r ∈ R, rx = x, hence r2x = rx and (r2 − r) x = 0. Let N = {y ∈ J | yx = 0}. Then N is a proper left ideal in J and therefore N is nilpotent. Thus, we obtain r2 ≡ r mod N. Let n = r2 − r, then (r + n − 2rn)2 ≡ r2 + 2rn − 4r2n mod N 2, r2 + 2rn − 4r2n ≡ r + n − 2rn mod N 2. 6 REPRESENTATION THEORY WEEK 9

2 Hence r1 = r + n − 2rn is an idempotent modulo N . Repeating this process several times we obtain an idempotent.  Corollary 4.2. If an artinian ring does not have nilpotent ideals, then it is semisim- ple. Proof. The sum S of all minimal left ideals is semisimple. By DCC S is a finite direct sum of minimal left ideals. Then S contains an idempotent e, which is the sum of idempotents in each direct summand. Then R = S ⊕ R(1 − e), however that implies R = S.  Important notion for a ring is the radical. For an R-module M let Ann M = {x ∈ R | xM = 0} . Then the radical rad R is the intersection of Ann M for all simple R-modules M. Theorem 4.3. If R is artinian then rad R is a maximal nilpotent ideal. Proof. First, let us show that rad R is nilpotent. Assume the contrary. Then rad R contains an idempotent e. But then e does not act trivially on a simple quotient of Re. Contradiction. Now let us show that any nilpotent ideal N lies in rad R. Let M be a simple module, then NM =6 M as N is nilpotent. But NM is a submodule of M. Therefore NM = 0. Hence N ⊂ Ann M for any simple M.  Corollary 4.4. An artinian ring R is semisimple iff rad R = 0. Corollary 4.5. If R is artinian, then R/ rad R is semisimple. Corollary 4.6. If R is artinian and M is an R-module, then for the filtration M ⊃ (rad R) M ⊃ (rad R)2 M ⊃···⊃ (rad R)k M = 0 all quotients are semisimple. In particular, M always has a simple quotient. Theorem 4.7. If R is artinian, then it has finite length as a left module over itself.

i Proof. Consider the filtration R = R0 ⊃ R1 ⊃··· ⊃ Rs = 0 where Ri = (rad R) . Then each quotient Ri/Ri+1 is semisimple of finite length. The statement follows.  Let R be an artinian ring. By Krull-Schmidt theorem R (as a left module over itself) has a decomposition into direct sum of indecomposable submodules R = L1 ⊕ op ···⊕ Ln. Since EndR (R) = R , the projector on each component Li is given by multiplication on the right by some idempotent ei. Thus, R = Re1 ⊕···⊕ Ren, where ei are idempotents and eiej = 0 if i =6 j. This decomposition is unique up to multiplication on some unit on the right. Since Rei is indecomposable, ei can not be written as a sum of two orthogonal idempotents, such idempotents are called primitive. Each module Rei is projective. REPRESENTATION THEORY WEEK 9 7

Lemma 4.8. Let R be artinian, N = rad R and e be a primitive idempotent. Then Ne is a unique maximal submodule of Re. Proof. Since Re is indecomposable, every proper left ideal is nilpotent, (otherwise it has an idempotent and therefore splits as a direct summand in Re). But then this ideal is in N ∩ Re = Ne.  A projective module P is a projective cover of M if there exists a surjection P → M. Theorem 4.9. Let R be artinian. Every simple R-modules S has a unique (up to an isomorphism) indecomposable projective cover isomorphic to Re for some primitive idempotent e ∈ R. Every indecomposable projective module has a unique (up to an isomorphism )simple quotient. Proof. Every simple S is a quotient of R, and therefore it is a quotient of some indecomposable projective P = Re. Let φ : P → S be the natural projection. For any indecomposable projective cover P1 of S with surjective morphism φ1 : P1 → S there exist f : P → P1 and g : P1 → P such that φ = φ1 ◦f and φ1 = φ◦g. Therefore φ = φ ◦ g ◦ f. Since Ker φ is the unique maximal submodule, g ◦ f (P1) = P . In particular g is surjective. Indecomposability of P1 implies P =∼ P1. Thus, every simple module has a unique indecomposable projective cover. On the other hand, let P be an indecomposable projective module. Corollary 4.6 implies that P has a simple quotient S. Hence P is isomorphic to the indecomposable projective cover of S.  Corollary 4.10. Every indecomposable projective module over an artinian ring R is isomorphic to Re for some primitive idempotent e ∈ R. There is a bijection between the ismorphism classes of simple R-modules and isomorphism classes of projective indecomposable R-modules.

Example. Let R = F3 (S3). Let r be a 3-cycle and s be a transposition. Then r − 1, r2 − 1, sr − s and sr2 − s span a maximal nilpotent ideal. Hence R has two (up to an isomorphism) simple modules L1 and L2, where L1 is a trivial representation of S3 and L2 is a sign representation. Choose primitive idempotents e1 = −s − 1 and e2 = s − 1, then 1 = e1 + e2. Hence R has two indecomposable projective modules P1 = Re1 and P2 = Re2. Note that ∼ S3 ∼ S3 Re1 = IndS2 (triv) , Re2 = IndS2 (sgn) .

Thus, P1 is just 3-dimensional permutation representation of S3, and P2 is obtained from P1 by tensoring with sgn. It is easy to see that P1 has a trivial submodule as well as a trivial quotient, and sgn is isomorphic to the quotient of the maximal submodule of P1 by the trivial submodule. One can easily get a similar description for P2 Thus, one has the the following exact sequences

0 → L2 → P2 → P1 → L1 → 0, 0 → L1 → P1 → P2 → L2 → 0, 8 REPRESENTATION THEORY WEEK 9 therefore ···→ P1 → P2 → P2 → P1 → P1 → P2 → P2 → P1 → 0 is a projective resolution for L1, and

···→ P2 → P1 → P1 → P2 → P2 → P1 → P1 → P2 → 0 is a projective resolution for L2. Now one can calculate Ext between simple modules k k Ext (Li, Li)=0if k ≡ 1, 2 mod 4, Ext (Li, Li)= F3 if k ≡ 0, 3 mod 4, and if i =6 j, then k k Ext (Li, Lj)=0if k ≡ 0, 3 mod 4, Ext (Li, Lj)= F3 if k ≡ 1, 2 mod 4.